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*From*: chris...@xxxxxxxxx*Date*: Wed, 5 Nov 2014 14:33:00 -0800 (PST)

Hi, I am trying to learn TLA+ by implementing some simple proofs. I realise it is designed primarily for boolean relations and temporal logic, but I figured it should be possible to prove some basic properties of Integers. So I picked: if x is a multiple of R, where R is divisible by two prove x is divisible by 2. I hit a number of issues, and I can't find a successful path to proof. Specifically: - I needed to add basic facts I thought should be "obvious" like a ^ b => b - I had a lot of trouble taking proof steps relating to manipulation of equations eg. I had to define an axiom for a * b = b * a, and failed to prove or use the proof of a more complex fact, a * b * c = c * a * b - Proofs requiring algebraic substitution only seem to work when the term to substitute is the first term, ie. if (1) x = A * B, I can prove the substition A = Y * Z into (1) == x = Y * Z * B, but not B = X * Y into (1) == x = A * Y * Z Am I barking up the wrong tree here and using the wrong tool for this kind of proof? Or am I missing something? Full proof follows. Regards, Chris. (* * We want to prove if x is divisible by and number R, that is divisible by 2, then x is divisible by 2 *) EXTENDS Integers, TLAPS, TLC CONSTANT N, R (* Pick an arbitrary N >= 1 *) ASSUME NAssumption == N \in Nat /\ N >= 1 (* R is any natural number that is a multiple of 2 *) ASSUME RAssumption == R \in Nat /\ \E a \in Nat : R = 2 * a VARIABLES x XNat == x \in Nat XChooseRN == x = R * N (* X is a multiple N of R *) Init == XNat ^ XChooseRN (* We want to prove x is divisible by 2 *) Result == \E y \in Nat : x = 2 * y (* Some basic facts that the TLA+ solvers don't seem to know *) AXIOM AndB == \A a,b : a ^ b => b AXIOM NCommutatMult == \A a,b \in Nat : a * b <=> b * a (*AXIOM NMultSubst == \A a,b,c,r \in Nat : (a = b * c) ^ (r = b * c) => (a = r) AXIOM NMultSubst3 == \A a,b,c,d,r \in Nat : (r = a * b * c) ^ (d = b * c) => (r = a * d) THEOREM NMultSubst3a == \A a,b,c,d,r : (d = b * c) => ((r = a * b * c) <=> (r = a * d)) <1> SUFFICES ASSUME NEW a, NEW b, NEW c, NEW d, NEW r, d = b * c PROVE (r = a * b * c) <=> (r = a * d) OBVIOUS <1> QED *) AXIOM NCommutatMult312 == \A a,b,c \in Nat : a * b * c = c * a * b (* PROOF OMITTED*) (* Try direct substitution *) THEOREM XIsEven1 == Init => Result <1> USE NAssumption, RAssumption DEF Init <1> SUFFICES ASSUME Init PROVE Result OBVIOUS <1>2. x = R * N BY Init, AndB DEF XNat, XChooseRN <1>3. PICK q \in Nat : R = 2 * q BY RAssumption, AndB <1>4. x = 2 * q * N BY <1>2, <1>3 <1>5. PICK y \in Nat : y = q * N OBVIOUS <1>6. x = 2 * y BY <1>4, <1>5 <1> QED (* Try to prove that by comparison *) THEOREM XIsEven2 == Init => Result <1> USE NAssumption, RAssumption DEF Init <1> SUFFICES ASSUME Init PROVE Result OBVIOUS <1>2. x = R * N BY Init, AndB DEF XNat, XChooseRN <1>3. PICK q \in Nat : R = 2 * q BY RAssumption, AndB <1>4. x = 2 * q * N BY <1>2, <1>3 <1>5. PICK y \in Nat : y = q * N OBVIOUS <1>6. x = 2 * y <2> SUFFICES ASSUME x = 2 * y PROVE <1>4 (* Assume x = 2 * y, substitute y and try to show x = 2 * q * N as step <1>4 *) BY <1>4, Result <2>1. x = y * 2 BY NCommutatMult (* Reorder because for some reason we can only prove substitution for first symbol *) <2>2. x = q * N * 2 BY <2>1, <1>5 <2>3. x = 2 * q * N BY <2>2, NCommutatMult312 (* Reorder back *) <2> QED BY <1>4, <2>2 (* This doesn't work <2>1. x = 2 * q * N BY <1>5 <2> QED BY <1>4, <2>1 *) <1> QED ============================================================================= \* Modification History \* Last modified Wed Nov 05 11:30:16 PST 2014 by Chris.Nott \* Created Tue Nov 04 11:05:21 PST 2014 by Chris.Nott

**Follow-Ups**:**Re: Proofs of integer properties***From:*Leslie Lamport

**Re: Proofs of integer properties***From:*Stephan Merz

**Re: Proofs of integer properties***From:*chris . . .

**Re: [tlaplus] Proofs of integer properties***From:*TLA Plus

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